CHAPTER 5 1

STRAIGHT-LINE (SL) METHOD

A constant amount is depreciated each year over the asset's life.

N = depreciable life of the asset in years.

dk = annual depreciation deduction in year k

dk = (B - SVN ) / N for k = 1, 2, ..., N

dk* = cumulative depreciation through year k.

dk* = k x dk

BVk = B - dk *

CHAPTER 5 2

DECLINING BALANCE (DB) METHOD

Annual depreciation is a constant percentage of the asset's value at the BOY

R = 2/N 200% declining balance

R = 1.5/N 150% declining balance

d1 = B x R

dk = B(1-R)k-1 (R) = BVk-1 (R)

dk* = B[1 - (1 - R)k ]

BVk = B(1 - R)k

BVN = B(1 - R)N

CHAPTER 5 3

SL and DB Example

The La Salle Bus Company has decided to purchase a new bus for $85,000, with a trade-in of their old bus. The old bus has a trade-in value of $10,000. The new bus will be kept for 10 years before being sold. Its estimated salvage value at that time is expected to be $5,000. Compute the following quantities using (a) the straight-line method, (b) the 200% declining balance method depreciation deduction in the first year and the fourth

year

cumulative depreciation through year four

book value at the end of the fourth year

CHAPTER 5 4

Cost basis:

B = $10,000 + $85,000 = $95,000

trade-in value cash-cost

Deduction amounts are fixed for SL:

10 to1 k for $9,000 10

5,000 - 95,000 d k

Deduction ratios are fixed for DB:

SV10

N

)BV 0.2 (d thus,, 0.2 10

2 R 1-kk

200% DB

CHAPTER 5 5

Straight Line Method (k = 1 to 4)

EOY, k dk BVk

0 0 95,000

1 9,000 86,000 = 95,000 – 9,000

2 9,000 77,000 = 86,000 – 9,000

3 9,000 68,000 = 77,000 – 9,000

4 9,000 59,000 = 68,000 – 9,000

d4 d1 = BV4 = 95,000 – 36,000 d4* = 4 x 9000 = 36,000

CHAPTER 5 6

Straight Line Method (k = 5 to 10)

EOY, k dk BVk

5 9,000 50,000 = 59,000 – 9,000

6 9,000 41,000 = 50,000 – 9,000

7 9,000 32,000 = 41,000 – 9,000

8 9,000 23,000 = 32,000 – 9,000

9 9,000 14,000 = 23,000 – 9,000

10 9,000 5,000 = 14,000 – 9,000

SV10 = B – N x dk

CHAPTER 5 7

200% Declining Balance Method

(k = 1 to 4)

EOY, k dk BVk

0 0 95,000

1 19,000 76,000

2 15,200 60,800

3 12,160 48,640

4 9,728 38,912

d1 = BV0 x R = B x 0.2

= 95,000 x 0.2 = 19,000

BV1 = BV0 – d1 = B – d1

= 95,000 – 19,000

d4 = 48,640 x 0.2 d4* = 19,000 + ... + 9,728

= 56,088

BV4 = B – d4*

CHAPTER 5 8

200% Declining Balance Method

(k = 5 to 10)

EOY, k dk BVk

5 7,782 31,130

6 6,226 24,904

7 4,981 19,923

8 3,985 15,938

9 3,188 12,750

10 2,550 10,200

BV10 = B – d10* = BV9 – d10

CHAPTER 5 9

SL vs. DB

CHAPTER 5 10

Consideration of Income Taxes in EE

Notation:

Rk = gross revenues in year k

Ek = operating expenses in year k plus interest paid on borrowed capital

dk = depreciation allowance for year k

t = effective income tax rate used for computing income taxes

Tk = income tax liability for year k

Income tax represents a significant

cash outflow that we cannot ignore

CHAPTER 5 11

General Tax Procedure

BTCF = Before tax cash flow = R – E

NIBT = Net income before tax = R – E – d

T = tax liability = t ( NIBT ) = t (R – E – d)

NIAT = Net income after tax = NIBT – T

= (R – E – d) – t (R – E – d) = (1 – t)(R – E – d)

ATCF = After tax cash flow = NIAT + d

ATCF = BTCF – t (R – E – d)

CHAPTER 5 12

General Tax Procedure - Example

You invested $113,028 on an asset with the depreciable life of 10 years. You can earn $30,000 per year from this investment for 10 years. Asset has a negligible or zero MV at the end of its useful life. Published income tax rate is 40% on annual taxable income (NIBT). Use after-tax MARR of 15% per year, and straight line depreciation method.

a) NIBT?

b) NIAT?

c) ATCF?

d) Is it profitable investment after taxes?

CHAPTER 5 13

Solution to (a), (b) and (c)

d = 113,028 / 10 = $11,303 (depreciation amount)

(+) Net income 30,000

(-) Deprecation 11,303

(a) NIBT $18,697 (R – E – d)

(-) Income Tax (0.4) 7,479

(b) NIAT $11,218 (1 – t)(R – E – d)

(+) Depreciation 11,303

(c) ATCF $22,521 (1 – t)(R – E – d)+d

CHAPTER 5 14

Solution to part (d)

AW (15%) = ATCF – 113,028 (A/P, 15%, 10)

= 22,521 – 22,521 = $0

CHAPTER 5 15

Typical Before-Tax Cash Flow Diagram:

Typical After-Tax Cash Flow Diagram:

After-Tax Cash Flow Analysis

CHAPTER 5 16

CHAPTER 5 17

After-tax MARR

To perform an after-tax evaluation of a project's after-tax cash flows, we must use an after-tax MARR.

trate, tax income effective - 1

MARRtax -After MARR tax - Before

0.12 0.4) - (1 0.2 MARR t- 1

MARR MARR AT

ATBT

Example: Suppose the before-tax MARR = 20% and t = 40%. What is the approximate after-tax MARR?

CHAPTER 5 18

ATCF Analysis – Example

Investment $10,000

Net Annual Receipts $4,000/yr

Study Period 4 years

Market Value at EOY 4 $5,000

After-tax MARR 15%

Effective income tax rate 40%

Depreciable recovery period 5 years

Is this a worthwhile investment after taxes? Use Straight Line Method for depreciation.

CHAPTER 5 19

Step 1: Find depreciation amounts

for the study period of 4 years:

5 to1 k for $2,000 5

0 - 10,000 d k

EOY, k dk BVk

0 --- 10000

1 2000 8000

2 2000 6000

3 2000 4000

4 2000 2000

CHAPTER 5 20

Step 2: Determine the ATCF with tax

rate of 40%:

Tk

EOY, k BTCFk dk TIk (t = 0.4) ATCFk

0 - 10,000 - 10,000

4b 5000 3000 -1200 3800

1 4000 2000 2000 -800 3200

2 4000 2000 2000 -800 3200

3 4000 2000 2000 -800 3200

4a 4000 2000 2000 -800 3200

MV4 – BV4 = 5000 – 2000

CHAPTER 5 21

Step 3: Use the ATCF to evaluate

this investment @ MARR = 15%:

After-tax cash flow diagram:

PW(15%) = – 10,000 + 3,200 (P/A, 15%, 3)

+ 7,000 (P/F, 15%, 4) = + $1,309

CHAPTER 5 22

Affordable Investment - Example

An asset purchased at the price of (I), and it is depreciated for 10 years, with straight line method. Estimated market value @ EOY 10 is zero, how much can you afford to purchase this asset when it produces annual $30,000 net income for 10 years. Effective income tax is 40% and MARRAT = 15%.

CHAPTER 5 23

Solve for I?

d = (I – 0) / 10

ATCF = 30000 – 0.4 (30000 – d)

AW(15%) = – I(A/P,15%,10)

+ 30000(1 – 0.4) + 0.4 (I/10)

AW(15%) = 0

I = 30000(1 – 0.4) / ((A/P,15%,10) – 0.04)

I = $113,028 0.1993

CHAPTER 5 24

Non-depreciable asset - Example Construction cost of the facility $600,000

Purchasing cost of land $550,000

Annual gross income $230,000

Operating expenses per year $30,000

Facility will be depreciated for 5 years, using 200% DB

MARRAT 12% Tax rate 40%

Is the investment worthwhile after taxes for the study period of 5 years?

Note: Land will be kept after the five-year operation! Facility has a market value of zero @EOY5

CHAPTER 5 25

200% DB R = 2/5 = 0.4

EOY, k dk BVk

cost basis for only

depreciable asset

0 --- 600,000

1 240,000 360,000

2 144,000 216,000

3 86,400 129,600

4 51,840 77,760

5 31,104 46,656

600,000 x 0.4

600,000 – 240,000

BV5 MV5 = 0

CHAPTER 5 26

ATCF analysis with t = - 0.4

ATCF

-1,150,000

216,000

177,600

154,560

140,736

132,442

Tax

16,000

-22,400

-45,440

-59,264

-67,558

TI

-40,000

56,000

113,600

148,160

168,896

d

240,000

144,000

86,400

51,840

31,104

EOY

0

1

2

3

4

5a

5b

BTCF

-1,150,000

200,000

200,000

200,000

200,000

200,000

550,000 - 46,656

MV5 – BV5 = 0 – 46,656

18,662

550,000 + 18,662

568,662

CHAPTER 5 27

ATCF analysis with MARRAT = 12%

ATCF

-1,150,000

216,000

177,600

154,560

140,736

132,442

568,662

EOY

0

1

2

3

4

5a

5b

+ (132,442 + 568,662)(P/F, 12%, 5)

= - $218,283 < 0 reject !

PW(12%) = - 1,150,000

+ 216,000(P/F, 12%, 1)

+ 177,600(P/F,12%, 2)

+ 154,560(P/F, 12%, 3)

+ 140,736(P/F, 12%, 4)

CHAPTER 5 28

Lease versus Purchase - Example Determine the more economic means of acquiring a

copier in your business if you may either:

a) purchase the copier for $5,000 with a probable resale value of $1,000 at the end of 5 years or

b) rent the copier for an annual fee of $900 per year for 5 years with an initial deposit of $500 refundable upon returning the copier in good condition.

If you own the copier, you will depreciate it by using the 150% DB method (class life of 5 years). All rental fees are deductible for income tax purposes. As the owner or lessee, you will pay all expenses associated with the operation of the copier. A deposit does not affect taxes when paid out or received back.

Compare these alternatives by using the equivalent uniform annual cost method. The after-tax MARR is 10% per year, and the effective income tax rate is 40%.

CHAPTER 5 29

Option A – Purchase Copier

EOY BTCF

R=0.3

d TI

t= -0.4

Tax ATCF

0 -5000 -5000

1 1500 -1500 600 600

2 1050 -1050 420 420

3 735 -735 294 294

4 515 -515 206 206

5a 360 -360 144 144

5b 1000 840 160 -64 936

CHAPTER 5 30

Option A – Purchase Copier EOY ATCF

0 -5000

1 600

2 420

3 294

4 206

5a 144

5b 936

AW(10%) = [-5000

+ 600(P/F,10%,1)

+420(P/F,10%,2)

+294(P/F,10%,3)

+206(P/F,10%,4)

+(144+936)(P/F,10%,5)](A/P,10%,5)

= - 3075(A/P,10%,5) = - 811.185

CHAPTER 5 31

Option B – Rent (lease) Copier

EOY BTCF

d TI

t = -0.4

Tax ATCF

0 -500 -500

1 -900 -900 360 -540

2 -900 -900 360 -540

3 -900 -900 360 -540

4 -900 -900 360 -540

5a -900 -900 360 -540

5b 500 500

CHAPTER 5 32

Option B – Rent (lease) Copier EOY ATCF

0 -500

1 -540

2 -540

3 -540

4 -540

5a -540

5b 500

AW (10%) = -500(A/P,10%,5)

+ 500(A/F,10%,5) – 540

= - 590

Option B (lease) is the least cost

alternative for having the coppier

CHAPTER 5 33

Before tax leasing cost? Over what range of before-tax leasing costs

would you choose the purchase option based on an after-tax analysis?

AW_lease =AW_purchase

AW_lease = - 811.185

-500(A/P,10%,5) + 500 (A/F,10%,5) + (1 – t) L

say L = before tax leasing cost

0.6 L = 811.185 + 500(A/P,10%,5) – 500 (A/F,10%,5)

L = 1269 (if before tax leasing cost is greater than 1269,

purchasing option will be preferred)